0.3 grams of platinum chloride of organic diacidic base left 0.09 grams of platinum on ignition,

the molecular weight of the organic base is  

$2B+H_{4}ptC{l}_6\to B_2{H}_{2}ptC{l}_6\stackrel{\Delta }{\to }pt$

Equivalent weight of $B_2{H}_{2}ptC{l}_6$

= 2B + 410

$\frac{0.3}{2B+410}=\frac{0.09}{195}$

$\Rightarrow$ 3.33 × 195 = 2B + 410

239.93 = 2B

B = 120

Molecular weight of base = 2 × E. Wt

                                                   = 2 × 120

                                                   = 240